[SOLVED] query from db, add values in dropdown menu

[jjmusicpro]

i wanted to query my db, and then add values to drop down menu, where when they select what they want in the dropdown, it will bring them to another page accounts.php where they will have more options, any help?

[BlueSkyIS]

1. Query the database to get the ids and/or values

2. Open a SELECT control: <SELECT name='something'>

3. Loop over the query results, inserting an OPTION for each one:

<OPTION value='{$row['id']}'>{$row['name']}</OPTION> or similar...

4. Close the select: </SELECT>

5. On POST, check the value of the control: $someval = $_POST['something'];

6. header to whatever page you want to go to: header("location: someotherpage.php");exit;

[jjmusicpro]

Do you have a code example?

 

I am new to php, thank you for all your help :D

[hamza]

Adding in any menu is quite simple.

      ...............................................................

      Just select the information or data from the database.

      Use the <select> tag and put the selected information

      into the <option> tag .

      And your menu will be created.

      its quite simple just try it .

      Reply me if you face any problem. ???

      .............................................................

[jjmusicpro]

I did try it, and gives error

[mattal999]

<form action='accounts.php' method='GET'>
<select name='name'>
<?php
parse_str("$QUERY_STRING");
$db = mysql_connect("[your host - usually "localhost"]", "[your username]","[your password]") or die("Could not connect.");
if(!$db) 
die("no db");
if(!mysql_select_db("[your database]",$db))
die("No database selected."); 
$acc1="SELECT * from accounts";
$acc2=mysql_query($acc1) or die("Could not select accounts.");
while($acc3=mysql_fetch_array($acc2)) {
echo "<option value='$query[name]'>$query[name]</option>";
}
?>
</select>
<input type='submit' name='submit' value='Submit'>
</form>

 

that would connect to a mysql database, select table accounts with one field, name, and print all the rows.

[jjmusicpro]

thank you very much, now i have a column named owner, and i have a owner listed multiple times, how do i make it only display unique results?

[jjmusicpro]

I tried to run that code and it didnt work

It builds the page, and the drop down, but nothings in the drop down

 

I have a table called zipcodes, with a column called account_name

 

<form action='nothing.php' method='GET'>
<select name='name'>
<?php
parse_str("$QUERY_STRING");
$db = mysql_connect("[localhost]", "[xxxxxxx]","[xxxxx]") or die("Could not connect.");
if(!$db) 
die("no db");
if(!mysql_select_db("[xxxxxx]",$db))
die("No database selected."); 
$acc1="SELECT * from zipcodes";
$acc2=mysql_query($acc1) or die("Could not select accounts.");
while($acc3=mysql_fetch_array($acc2)) {
echo "<option value='$query[account_name]'>$query[account_name]</option>";
}
?>
</select>
<input type='submit' name='submit' value='Submit'>
</form>

[jjmusicpro]

Anyone? 

[marcus]

$sql = "SELECT * FROM `table`";
$res = mysql_query($sql) or die(mysql_error());

if(mysql_num_rows($res) == 0){
echo "No values in database!\n";
}else {
echo "<select name=\"zip\">\n";
   while($row = mysql_fetch_assoc($res)){
   echo "<option value=\"$row[zip]\">$row[town]</option>\n";
   }
echo "</select>\n";
}

[jjmusicpro]

I tried that but didnt work:

Still gives blank drop down...

 

<form action='nothing.php' method='GET'>
<select name='name'>
<?php
parse_str("$QUERY_STRING");
$db = mysql_connect("[localhost]", "[xxxxx]","[xxxxxxx]") or die("Could not connect.");
if(!$db) 
die("no db");
if(!mysql_select_db("[xxxxxxx]",$db))
die("No database selected."); 
$sql = "SELECT * FROM `zipcodes`";
$res = mysql_query($sql) or die(mysql_error());

if(mysql_num_rows($res) == 0){
echo "No values in database!\n";
}else {
echo "<select name=\"zip\">\n";
   while($row = mysql_fetch_assoc($res)){
   echo "<option value=\"$row[account_name]\">$row[iD]</option>\n";
   }
echo "</select>\n";
}

?>
</select>
<input type='submit' name='submit' value='Submit'>
</form>

[marcus]

Is $row[iD] in caps?

[jjmusicpro]

I changed it but sill didnt want to work: The page comes up with a dropdown, but nothing is in the drop down:

 

<form action='nothing.php' method='GET'>Select Account:<br>
<select name='name'>
<?php
parse_str("$QUERY_STRING");
$db = mysql_connect("[localhost]", "[xxxxxx]","[xxxxx]") or die("Could not connect.");
if(!$db) 
die("no db");
if(!mysql_select_db("xxxxxxx]",$db))
die("No database selected."); 
$sql = "SELECT * FROM `zipcodes`";
$res = mysql_query($sql) or die(mysql_error());

if(mysql_num_rows($res) == 0){
echo "No values in database!\n";
}else {
echo "<select name=\"zip\">\n";
   while($row = mysql_fetch_assoc($res)){
   echo "<option value=\"$row[account_name]\">$row[search_id]</option>\n";
   }
echo "</select>\n";
}

?>
</select>
<input type='submit' name='submit' value='Submit'>
</form>

[mattal999]

<form action='nothing.php' method='GET'>
<select name='name'>
<?php
parse_str("$QUERY_STRING");
$db = mysql_connect("localhost", "username","password") or die("Could not connect.");
if(!$db) 
die("no db");
if(!mysql_select_db("dbname",$db))
die("No database selected."); 
$acc1="SELECT * from zipcodes";
$acc2=mysql_query($acc1) or die("Could not select accounts.");
while($acc3=mysql_fetch_array($acc2)) {
echo "<option value='$acc3[account_name]'>$acc3[search_id]</option>"; // was $query[account_name] - not set, i set $acc3 - this was my old one
}
?>
</select>
<input type='submit' name='submit' value='Submit'>
</form>

[jjmusicpro]

Thank you very much, how do i limit the results to just 1 instance of that account_name found?

 

Right now, i might have 4 or 5 account_name listed the same, and it will put all 5 of those in the drop down, i just want 1 of each of the uniquie account_name it finds

[jjmusicpro]

I got it i just added this:

 

$acc1="SELECT account_name from zipcodes group by account_name";

 

Thanks again!